and real rates is given by the Fisher equation, which is This states that the real interest rate () equals the nominal interest rate () minus the expected inflation rate (). Here all the rates are continuously compounded. For rates based on simple interest, the Fisher equation takes the form where is the simple nominal interest rate and is the simple real interest rate; this equation is well approximated by using the simple rates in the previous equation provided all three percentage rates
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the syllabus and overall progression of our study of functions and relationships of variables, this week we took a look at Two-Variable inequalities. The assignment was a real world application of using these types of equations to solve shipping problems. We can use these equations to figure out the solutions to common problems and provide graphs to show a range of correct and incorrect answers that is easier to read. The problem that was given, #68 on pg. 539 shows a graph that illustrates how
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| 2.91 x 10-6 | 0.00175 | -12.75 | 673 | 8.38 x 10-4 | 0.00149 | -7.08 | 773 | 7.65 x 10-2 | 0.00130 | -2.57 | Plot of ln k against 1/T (K-1) Slope of graph, m = -7.08-(-12.75)0.00149-0.00175 = 5.67- 0.00026 = 21807.69 Using Arrhenius equation, m = - Ea8.314 J/Kmol 21807.69 = - Ea8.314 J/Kmol Ea = 181 kJ/mol Question 4 (a) Rate of reaction of NO is second order. Rate of reaction of H2 is first order. Total rate of reaction = third order. (b) If [NO] is doubled, the
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101) SYLLABUS I. Basic algebra: Linear and quadratic equations, Solving linear and quadratic equations, Application of equations: profit, pricing, savings, revenue, sales tax, investment, bond redemption, linear inequalities, applications of inequalities: profit, renting verses purchasing, leasing versus purchasing, revenue, current ratio, investment, Maple session on solving linear, quadratic and higher degree equations, solving inequalities II. Functions and Graphs: Introduction
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Identification Status: Equation 1: P1=1, P2=1 so that P1=P2 so, equation is Exactly identified. Equation 2: P1=0, P2=1 so that P1<P2 so, equation is Unidentified. ii. Reduced form equations: Putting Y1 in Y2: Y2= β0+β1(∝0+∝1Y2+∝2X1+∈1) +β2X1+β3X3+ϵ2 Y2= β0+β1α0+β1α1Y2+β1α2X1+β1ϵ1+β2X1+β3X3+ϵ2 Y21-β1α1= β0+β1α0+X1β1α2+β2+β3X3+β1ϵ1+ϵ2 Y2= β0+β1α01-β1α1+β1α2+β21-β1α1X1+β31-β1α1X3+β1ϵ1+ϵ21-β1α1 Y1=π20+π21X1+π22X3+ν2 Now putting this reduced form equation of Y2 in Y1 equation: Y1= ∝0+∝1(π20+π21X1+π22X3+V2)∝2X1+∈1
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Contents I. Abstract: 2 II. Question 1 3 III. Question 2 3 A. Normal Incidence 3 B. Oblique Incidence 5 1. Perpendicular Polarization 5 2. Parallel Polarization 12 IV. Question 3: 17 A. Cost of conducting Material 18 B. Mass of conducting Material 18 C. Volume of Conducting Material 19 V. Conclusion: 19 Abstract: The purpose of this project is to design a shielded room. The room should be built in a way so that high-frequency waves cannot penetrate into the room. The material that
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of varying strengths and needs, I often get the question "when are we ever going to use this stuff?" It's hard to explain to a 14 year old that just because she will not write academic papers in her work life or because he won't use the quadratic equation when going about his home life, doesn't mean it isn't helping to develop the student's brain and cognitive structures. Helping students to understand the difference between obliterative subsumption and forgetting may help motivate them, even a little
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y = mx + b 2. Quadratic Functions – it has a degree and forms a parabolic path. The highest (or lowest point) of the parabola is called the vertex. At has a form of (standard form of quadratic equation) F(x) = Ax2 + Bx + C where A, B,C are constant. Vertex form of Quadratic F(x) = a (x-h)2 + K Quadratic Formula 3. Polynomial Functions – a quadratic, a cubic, a quartic and so on involving only non-negative
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know P(0) = 0, P'(0) = 0, P(L) = h, P'(L) = 0 The original equation is P(x)= ax3 + bx2 + cx + d P(0) = 0, so plug in zero for x. You will find that d = 0. Simplify the equation to P(x)= ax3 + bx2 + cx Find the derivative of P(X), which is 3ax2 + 2bx + c P’(0) = 0, so plug in zero for x. We found that c =0 Simplify the equation to P’(X) = 3ax2 + 2bx Plug in P'(L) = 0 We get the equation of 3aL2+ 2bL= 0 ① Plug in P(L) = h We get the equation of aL3 + bL2 = h ② Solve for a,b 3aL2+ 2bL= 0
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AND SECTION: 3A, 3B, 3C, 3D CREDIT : 3 Units SCHOOL YEAR: 1ST SEMESTER 2013-2014 COURSE DESCRIPTION: This course is designed for Commerce and Accountancy students which deals with the basic algebraic expressions, linear equations and inequalities, solving problems in a linear programming involving graphical method, simplex method, transportation method and assignment method, the break even analysis, the decision theory, business forecasting and inventory. GRADING SYSTEM:
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